Problem:
Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including 1 and itself. Find n/75.
Solution:
Suppose the prime factorization of n has the form
n=p1r1p2r2⋯pkrk
where p1,p2,…,pk are the distinct prime divisors of n and r1,r2,…,rk are positive integers. Then the number of divisors of n is given by
(r1+1)(r2+1)⋯(rk+1)
Since this last product must be 75=3⋅5⋅5, we see that n can have at most three distinct prime factors. To ensure that n is divisible by 75 and that the n we obtain is minimal, the prime factors must belong to the set {2,3,5}, with the factor 3 occurring at least once and the factor 5 occurring at least twice. Thus
n=2r13r25r3
with
(r1+1)(r2+1)(r3+1)=75r2≥1,r3≥2
It is not hard to write down the ordered triples (r1,r2,r3) that satisfy the above conditions:
(4,4,2)(4,2,4)(2,4,4)(0,4,14)(0,14,4)(0,2,24)(0,24,2)
Among the above ordered triples, the minimum value for n occurs when r1=r2=4 and r3=2. Thus our answer is 75n=2433=432.
The problems on this page are the property of the MAA's American Mathematics Competitions